RC Example 3

RC IV

For more info., please contact: Eng. MHD. [email protected]

AIU 2014-2Arab International University

EXAMPLE 03

Example 3

E.3

Figure.1

Figure.2

Given

Building: Five-story Residential BuildingPlace: SaydnayaLateral Force-Resisting System: Shear WallsBearing Capacity of the Soil: 3.5kg/cm2Live Load: 2kN/m2Total Dead Load Typical Floor: 12kN/m2Total Dead Load Last Floor: 8kN/m2Story Height: 3.25mShear Walls - Thickness: 25cm

Required

Base ShearVertical Distribution of the ForceStory Shears Overturning and Stabilizing MomentsTotal Shear Force Affecting Each Shear Wall

Solution 1 Base Shear, Vy


Cv = 0.2


Cv = 0.2

I = 1


hn = 3.25x5 = 16.25 m

T = 0.3950 Sec


Cv = 0.2

I = 1

T = 0.3950


Cv = 0.2

I = 1

T = 0.3950

R = 4.5


3.25m

3.25/2

3.25/2

Typical StoryW = W(kN/m2) x A(m2) = 12 x (20x12) = 2880kN

Last StoryW = W(kN/m2) x A(m2) = 8 x (20x12) = 1920kN

All StoriesW = 4x2880 + 1920 = 13440kN


Cv = 0.20

I = 1

T = 0.3950

R = 4.5

W = 13440kN

V = 1512.2kN

Vmax = 1493.3

V = 1512.2

Vmin = 295.7

V = 1512.2

V = 1493.3kN

Solution 2 Vertical Distribution

T = 0.3950

V = 1493.3

Ft = 41.3kN

0.25 x 1493.3 = 373.3

Ft = 41.3kN

T= 0.3950 < 0.7

Ft = Zero


Ft = Zero

V = 1493.3

Storyh (m)wi (kN)wi . hiFxStory Shear

516.25192031200373.33373.33

413288037440447.99821.32

39.75288028080335.991157.31

26.5288018720224.001381.30

13.2528809360112.001493.30

Sum134401248001493.30

Solution 3 S.F.D Story Shears

Solution 3 Overturning


StoryFxOverturning Moments

5373.33M5 = 373.33x3.25 = 1213.32

4447.99M4 = 373.33x6.5 + 447.99x3.25 = 3882.61

3335.99M3 = 7643.87

2224.00M2 = 12133.13

1112.00M1 = 16986.39


Mb = 16986.4 kN.m

Mf = W . (b/2)

Mf / Mb = 4.75

Mf = 80640 kN.m

1.5

5 W1: Direct Portion, F

F = (Ixi/SIxi) Fy

F = (1.333/(3x1.333)) Fy

F = (1/3) Fy

Fy

F1

F2

F3

Ii = (ti x Li3)/12

Ix = (0.25 x 43)/12 = 1.333m4

Iy = (0.25 x 53)/12 = 2.604m4


StoryFyMtF = (1/3)FyFF

5373.33124.44

4447.99149.33

3335.99112.00

2224.0074.67

1112.0037.33

5 Center of Mass&Center of Rigidity

Symmetrical in Both; Mass & Rigidity??

5 Center of Mass

5 Center of Rigidity

xR = S(Ixi . xi) /SIxi

xR = 1.333(0.125)+1.333(5.125)+1.333(19.875) / (3x1.333)

xR = 8.375m

yR = S(Iyi . yi) /SIyi

YR = 2.604(0.125)+2.604(11.875) / (2x2.604)

yR = 6m

5 Actual Eccentricity, ex , ey

ex = xM xR = 10 8.375 = 1.625m

ey = yM yR = 6 6 = 0

5 Design Eccentricity, exd , eyd

eyd = ey 0.05 x b

eyd = 0 0.05 x 12

exd = ex 0.05 x a

exd = 1.625 0.05 x 20

exd = 2.625 mexd = 0.625 m

eyd = + 0.6 meyd = - 0.6 m

5 Torsional Moment, Mt

Fy

F3

F2

F1

Mt


Mt = Fy x exd

Mt = 2.625 FyMt = 0.625 Fy

Mt = Fy x 2.625

Mt = Fy x 0.625

StoryFyMt = 2.625 FyMt = 0.625 Fy

5373.33979.98233.33

4447.991175.97279.99

3335.99881.98210.00

2224.00587.99140.00

1112.00293.9970.00

5 W1: Torsional Portion, F

F = (Ixi . xi / IP) Mt

F1

5 Rotational Stiffness, IP

IP = 1.333x8.252 + 1.333x3.252 + 1.333x11.52 + 2x(2.604x5.7852)

IP = 455.39 m6

IP = S Ixi(xi)2 + S Iyi(yi)2



F = (1.333x(-8.25) / 455.39) Mt

F = (- 0.024) Mt

F1

Mt = 2.625 FyMt = 0.625 Fy

Min of


StoryFyMt = 0.625 FyFF = -0.024MtF

5373.33233.33-5.60

4447.99279.99-6.72

3335.99210.00-5.04

2224.00140.00-3.36

1112.0070.00-1.68

5 W1: Total Shear Force, F = F + F

StoryFyMtFFF

5373.33233.33124.44-5.60118.84

4447.99279.99149.33-6.72142.61

3335.99210.00112.00-5.04106.96

2224.00140.0074.67-3.3671.31

1112.0070.0037.33-1.6835.65


F = (Ixi/SIxi) Fy

F = (1.333/(3x1.333)) Fy

F = (1/3) Fy

Fy

F1

F2

F3

Ii = (ti x Li3)/12

Ix = (0.25 x 43)/12 = 1.333m4



5373.33124.44

4447.99149.33

3335.99112.00

2224.0074.67

1112.0037.33



F = (1.333x(-3.25) / 455.39) Mt

F = (- 0.0095) Mt

F2

Mt = 2.625 FyMt = 0.625 Fy

Min of


StoryFyMt = 0.625 FyFF = -0.0095MtF

5373.33233.33-2.22

4447.99279.99-2.66

3335.99210.00-1.99

2224.00140.00-1.33

1112.0070.00-0.66


StoryFyMtFFF

5373.33233.33124.44-2.22122.23

4447.99279.99149.33-2.66146.67

3335.99210.00112.00-1.99110.00

2224.00140.0074.67-1.3373.34

1112.0070.0037.33-0.6636.67


F = (Ixi/SIxi) Fy

F = (1.333/(3x1.333)) Fy

F = (1/3) Fy

Fy

F1

F2

F3

Ii = (ti x Li3)/12

Ix = (0.25 x 43)/12 = 1.333m4



5373.33124.44

4447.99149.33

3335.99112.00

2224.0074.67

1112.0037.33



F = (1.333x(11.5) / 455.39) Mt

F = (+ 0.034) Mt

Mt = 2.625 FyMt = 0.625 Fy

Max of

F3


StoryFyMt = 2.625 FyFF = +0.034MtF

5373.33979.9833.32

4447.991175.9739.98

3335.99881.9829.99

2224.00587.9919.99

1112.00293.9910.00


StoryFyMtFFF

5373.33979.98124.4433.32157.76

4447.991175.97149.3339.98189.31

3335.99881.98112.0029.99141.98

2224.00587.9974.6719.9994.66

1112.00293.9937.3310.0047.33


F = (Iyi/SIyi) Fx

F = (2.604/(2x2.604)) Fx

F = (1/2) Fx

F5

F4

Ii = (ti x Li3)/12

Iy = (0.25 x 53)/12 = 2.604m4


StoryFxMtF = (1/2)FxFF

5373.33186.66

4447.99224.00

3335.99168.00

2224.00112.00

1112.0056.00


Mt = Fx x eyd

Mt = 0.6 FxMt = - 0.6 Fx

Mt = Fx x 0.6

Mt = Fx x (-0.6)

StoryFxMt = 0.6 FxMt = - 0.6 Fx

5373.33224.00-224.00

4447.99268.79-268.79

3335.99201.60-201.60

2224.00134.40-134.40

1112.0067.20-67.20


F = (Iyi . yi / IP) Mt

F = (2.604x(-5.875) / 455.39) Mt

F = (- 0.034) Mt

Mt = 0.6 FxMt = - 0.6 Fx

F4


StoryFxMt = - 0.6 FxFF = -0.034MtF

5373.33-224.007.62

4447.99-268.799.14

3335.99-201.606.85

2224.00-134.404.57

1112.00-67.202.28


StoryFyMtFFF

5373.33-224.00186.667.62194.28

4447.99-268.79224.009.14233.13

3335.99-201.60168.006.85174.85

2224.00-134.40112.004.57116.57

1112.00-67.2056.002.2858.28


F = (Iyi/SIyi) Fx

F = (2.604/(2x2.604)) Fx

F = (1/2) Fx

F5

F4

Ii = (ti x Li3)/12

Iy = (0.25 x 53)/12 = 2.604m4


StoryFxMtF = (1/2)FxFF

5373.33186.66

4447.99224.00

3335.99168.00

2224.00112.00

1112.0056.00


F = (Iyi . yi / IP) Mt

F = (2.604x(+5.875) / 455.39) Mt

F = (+ 0.034) Mt

Mt = 0.6 FxMt = - 0.6 Fx

F5


StoryFxMt = 0.6 FxFF = 0.034MtF

5373.33224.007.62

4447.99268.799.14

3335.99201.606.85

2224.00134.404.57

1112.0067.202.28


StoryFyMtFFF

5373.33-224.00186.667.62194.28

4447.99-268.79224.009.14233.13

3335.99-201.60168.006.85174.85

2224.00-134.40112.004.57116.57

1112.00-67.2056.002.2858.28

END OF PR. SESSION 05

TO BE CONTINUED NEXT SESSION

For more info., please contact: Eng. MHD. [email protected]

AIU 2014-2Arab International University

RC IV

EQUIVALENT STATIC METHOD

Documents

RC Example 3